Evaluate lim x->0 (Sin^3 3x - 3x^3/x^3 x^4)
I got zero (0) to be my answer by imputing 0 in place of x but am not so sure if the answer should be 0, am thinking it should be like 1/0 or something like that.
what does "x^3 x^4" mean?
lim x->0 (Sin^3 3x - 3x^3)/(x^3) (x^4)
To evaluate this limit, let's start by simplifying the expression:
lim[x->0] [(sin^3(3x) - 3x^3) / (x^3 * x^4)]
Next, we can use the following trigonometric identity: sin^3(x) = (3sin(x) - sin(3x)) / 4.
lim[x->0] [((3sin(3x) - sin(9x))/4 - 3x^3) / (x^3 * x^4)]
Now, let's distribute the x^3 * x^4 to each term:
lim[x->0] [(3sin(3x) - sin(9x) - 12x^3) / (4x^7)]
Now, we can evaluate the limit as x approaches 0.
When you substitute 0 for x in the expression, you will get:
[(3sin(0) - sin(0) - 12 * 0^3) / (4 * 0^7)]
= (0 - 0 - 0) / 0
At this point, we notice that the numerator is 0, but the denominator is also 0. This indicates an indeterminate form.
To further evaluate this limit, we can simplify the expression using L'Hopital's Rule. This rule allows us to take the derivative of both the numerator and the denominator, and then evaluate the limit again:
lim[x->0] [(9cos(3x) - 9cos(9x) - 36x^2) / (28x^6)]
Now, let's substitute 0 for x again:
[(9cos(0) - 9cos(0) - 36 * 0^2) / (28 * 0^6)]
= (9 - 9 - 0) / 0
We can see that we still have an indeterminate form. So, let's apply L'Hopital's Rule again:
lim[x->0] [(-27sin(3x) + 81sin(9x) - 72x) / (168x^5)]
Now, substitute 0 for x:
[(-27sin(0) + 81sin(0) - 72 * 0) / (168 * 0^5)]
= (0 + 0 - 0) / 0
Again, we have an indeterminate form. Applying L'Hopital's Rule one more time:
lim[x->0] [(-81cos(3x) + 729cos(9x) - 72) / (840x^4)]
And substituting 0 for x:
[(-81cos(0) + 729cos(0) - 72) / (840 * 0^4)]
= (-81 + 729 - 72) / 0
Now, we have an expression in which the numerator is a constant and the denominator is 0. This indicates that the limit is undefined.
Therefore, the answer to the limit is undefined rather than being equal to zero.